The truncation method for a
two-dimensional nonhomogeneous
backward heat problem
Phan Thanh Nam1 , Dang Duc Trong2 and Nguyen Huy Tuan3∗
1 Department
2 Faculty
of Mathematical Sciences, University of Copenhagen, Denmark.
of Mathematics and Computer Sciences, University of Science,
Vietnam National University, HoChiMinh city, Vietnam.
3 Faculty
of Mathematics , SaiGon University, HoChiMinh city, Vietnam.
Abstract
We consider the backward heat problem
ut − uxx − uyy = f (x, y, t), (x, y, t) ∈ Ω × (0, T ),
u(x, y, T ) = g(x, y), (x, y) ∈ Ω,
with the homogeneous Dirichlet condition on the rectangle Ω = (0, π) × (0, π),
where the data f and g are given approximately. The problem is severely
ill-posed. Using the truncation method for Fourier series we propose a simple
regularized solution which not only works on a very weak condition on the exact
data but also attains, due to the smoothness of the exact solution, explicit er√
ror estimates which include the approximation (ln(−1 ))3/2 in H 2 (Ω). Some
numerical examples are given to illuminate the effect of our method.
Keywords and phrases: Backward heat problem, Ill-posed problem, Nonhomogeneous heat, truncation method, error estimate.
Mathematics subject Classification 2000: 35K05, 35K99, 47J06, 47H10.
∗
Corresponding author. E-mail: tuanhuy [email protected]
1
1
Introduction
Let Ω = (0, 1) × (0, 1) be a heat conduction rectangle. Given the heat source f (x, y, t)
on (x, y, t) ∈ Ω × [0, T ] and the final temperature u(x, y, T ) at some time T > 0, we
consider the problem of recovering the temperature distribution u(x, y, t) from the
backward heat problem
ut − uxx − uyy = f (x, y, t), (x, y, t) ∈ Ω × (0, T ),
(1)
u(0, y, t) = u(π, y, t) = u(x, 0, t) = u(x, π, t) = 0, t ∈ (0, T ),
(2)
u(x, y, T ) = g(x, y), (x, y) ∈ Ω.
(3)
Since f and g come from measurement, they are in general non-smooth and only approximate values. This is a typical example of the inverse and ill-posed problem since
although this problem has at most one solution (see Theorem 1 in Section 2), the
solution does not always exist, and in the case of existence, it does not depend continuously on the given data. The instability makes the numerically calculus difficult
and hence a regularization is in order.
The homogeneous backward heat problems, i.e. the case f = 0, was extensively
considered by many authors using many approach, e.g. the original quasi-reversibility
method of Lattes and Lions [10], the quasi-boundary value problem method [15], the
quasi-solution method of Tikhonov [16], the logarithmic convexity method [1] and
the C-regularized semi-groups technique [7]. Physically, this problem arises from the
requirement of recovering the heat temperature at some earlier time using the knowledge about the final temperature. The problem is also involved to the situation of
a particle moving in a environment with constant diffusion coefficient (see [6]) when
one asks to determine the particle position history from its current place. The interest of backward heat equations also comes from financial mathematics, where the
celebrated Black-Scholes model [2] for call option can be transformed into a backward parabolic equation whose form is related closely to backward heat equations.
Although there are many papers on the homogeneous backward heat equation, the
result on the inhomogeneous case is very scarce while the inhomogeneous case is, of
course, more general and nearer to practical application than the homogeneous one.
Shortly, it allows the appearance of some heat source which is inevitable in nature.
Let us mention here some approachs and their technical difficulties of many earlier works. In the method of quasi-reversibility, the main ideas is of replacing the
2
unbounded operator A (in our case is −∆) by a perturbed one A . In the original
method in 1967, Lattes and Lions [25] proposed A (A) = A − A∗ A, i.e. adding a
”corrector” into the original operator, to obtain a well-posed problem. The essential
difficulty of the quasi-reversibility method is due to the appearance of the second-order
operator A∗ A which produces serious difficulties on the numerical implementation.
In addition, the stability magnitude of the approximating problem, i.e. the error inT
troduced by a small change in the final value, is of order e which is very large when
becomes small.
In 1983, Showalter [15] presented the quasi-boundary value method for the homogeneous problem which gave a stability estimate better than the one of the quasireversibility method discussed above. The main ideas of this method is of adding
an appropriate ”corrector” into the final data (instead of the main equation). Using
this method, Clark and Oppenheimer [3], and very recently Denche and Bessila [4],
regularized the backward heat problem by replacing the final condition by
u(T ) + u(0) = g
and
u(T ) − u0 (0) = g,
respectively. This method, in general, gives the stability estimate of order −1 .
Although there are many papers on the homogeneous case of the backward problem, we only find a few result on the inhomogeneous case, and especially the two
dimensional case is very scarce. In 2006, Trong and Tuan [17] approximated a one
dimensional inhomogeneous linear problem by the quasi-reversibility method. As we
T
mention before, the stability magnitude of the method is of order e . In their work
the error between the approximate problem and the exact solution is
r
8
∂ 4 f (x, t) 2
kL2 (0,T ;L2 (0,π)) ,
(T − t) 4 ku(., 0)k2 + t2 k
t
∂x4
which is very large when t becomes small. In 2007, Trong et al. [19] used an improved
version of quasi boundary value method to regularize the one-dimensional version of
(1-3) for a nonlinear heat source f = f (x, t, u). Their error estimate is t/T for t > 0
and (ln(1/))1/4 for t = 0.
One of the essential requirements of the previous works on inhomogeneous prob-
3
lem, e.g. [17, 19], is
∞
X
2
e2T k gk2 < ∞
(4)
k=1
where gk is the coefficient of the Fourier series of the final datum u(., T ) = g, i.e.
Z
2 π
gk =
g(x) sin(kx)dx.
π 0
While such a condition is reasonable in homogeneous problems, it is not necessarily
true in the inhomogeneous case. For example, consider the problem
ut − uxx = f (x, t) ≡ et x, (x, t) ∈ (0, π) × (0, T ),
u(0, t) = u(π, t) = 0, t ∈ (0, T ).
Corresponding to the final value u(x, 1) = g(x) ≡ ex, the equation has a (unique)
k+1
solution u(x, t) = et x. However, by direct computation we find that gk = 2e (−1)k
and hence
∞
X
2
e2T k gk2
2
= 4e
k=1
2
∞
X
e2T k
k=1
k2
> ∞.
In the present paper, we do not need condition (4). In fact, we shall give a simple
and convenient way to construct the regularized method which works with very weak
assumption on the exact solution.
Let us give a simple analysis for the ill-posedness of the problem(1-3). This problem may be rewritten formally as


ZT
∞
X
2
2
2
2
u(x, y, t) =
e(T −t)(m +n ) gmn − e(s−T )(m +n ) fmn (s)ds sin(mx) sin(ny) (5)
m,n=1
t
where gmn and fnm (t) are the coefficient of the Fourier-sin expansion of g and f (., ., t),
i.e.
Z
4
gmn :=
g(x, y) sin(mx) sin(ny)dxdy,
π2 Ω
Z
4
fmn (t) :=
f (x, y, t) sin(mx) sin(ny)dxdy.
π2 Ω
2 +n2 )
If t < T then e(T −t)(m
term e−(t−T )(m
2 +n2 )
increases very fast when m2 + n2 becomes large. Thus the
is the source of instability.
4
It is a natural think to recover the stability of problem (5) is to filter all high
frequencies. In the present paper, we simply do that by using the truncated regularization method, namely taking the sum (5) only for m2 +n2 ≤ M with an appropriate
regularization parameter M . The truncated regularization method is a very simple
and effective method for solving some ill-posed problems and it has been successfully
applied to some inverse heat conduction problems [5, 8, 13]. However, in many earlier
works, we find that only logarithmic type estimates in L2 -norm are available; and
estimates of Hölder type are very rate (see Remark 5 and Remark 6 for more detail
comparisons). In our method, corresponding to different levels of the smoothness of
the exact solution, the convergence rates will be improved gradually. In particular, if
√
we impose a condition similar to (4) then the error estimate in H 2 (Ω) is (ln())3/2 ,
which is better than any Hölder estimate of order q with q ∈ (0, 1/2). We mention
that our regularized solution in all case is unique, and all error estimates are valid for
all t ∈ [0, T ].
The remainder of the paper is organized as follows. In Section 2 we shall construct
the regularized and show that it works even with very weak condition on the exact
solution. In Section 3, many error estimates are derived, in both of the usual cases
such as the exact solution u in H01 (Ω) or H 2 (Ω), and the special cases when the
exact solution is very smooth. Some numerical experiments are given in Section 4 to
illuminate the effect of our method.
2
Regularized solution
Let us first make clear what a weak solution of the problem (1-3) is. As follows
we shall write u(t) = u(., ., t) for short. We call a function u ∈ C ([0, T ]; L2 (Ω)) ∩
C 1 ((0, T ); L2 (Ω)) to be a weak solution for the problem (1-3) if
d
hu(t), W iL2 (Ω) − hu(t), ∆W iL2 (Ω) = hu(t), ∆W iL2 (Ω) ,
dt
(6)
for all function W (x, y) ∈ H 2 (Ω) ∩ H01 (Ω). In fact, it is enough to choose W in the
orthogonal basis {sin(mx) sin(ny)}m,n≥1 and the formula (6) reduces to
(T −t)(m2 +n2 )
umn (t) = e
ZT
gmn −
2 +n2 )
e(s−t)(m
t
which may also be written formally as (5).
5
fmn (s)ds,
∀m, n ≥ 1
(7)
Note that if the exact solution u is smooth then the exact data (f, g) is smooth
also. However, the real data, which come from practical measure, is often discrete
and non-smooth. We shall therefore always assume that f ∈ L1 (0, T ; L1 (Ω)) and
g ∈ L1 (Ω), and the error of the data is given on L1 only. Note that (7) still makes
sense with such data, and this formula gives immediately the uniqueness.
Theorem 1 (Uniqueness). For each f ∈ L1 (0, T ; L1 (Ω)) and g ∈ L1 (Ω), the problem
(1 − 3) has at most one (weak) solution u ∈ C ([0, T ]; L2 (Ω)) ∩ C 1 ((0, T ); L2 (Ω)).
In spite of the uniqueness, the problem is still ill-posed and a regularization is
necessary. For each > 0, introduce the truncation mapping P : L1 (Ω) → C ∞ (Ω) ∩
H01 (Ω)
P w(x, y) =
X
m,n≥1;m2 +n2 ≤M
wmn sin(mx) sin(ny), with M =
ln(−1 )
.
2T
(8)
In fact, P is a finite-dimensional orthogonal projection on L2 (Ω), but it works on
L1 (Ω) as well. We shall approximate the original problem by the following well-posed
problem.
Theorem 2 (Well-posed problem). For each f ∈ L1 (0, T ; L1 (Ω)) and g ∈ L1 (Ω), let
w ∈ L1 (0, T ; L2 (Ω)) defined by
wmn (t) = e
(T −t)(m2 +n2 )
ZT
(P g)mn −
2 +n2 )
e(s−t)(m
(P f )mn (s)ds, ∀m, n ≥ 1.
(9)
t
Then w = P w and it depends continuously on (f, g), i.e. if wi is the solution with
respect to (fi , gi ), i = 1, 2, then
p
2 ln(ε−1 ) t−T √
ε 2T kg1 − g2 kL1 (Ω) + kf1 − f2 kL1 (0,T ;L1 (Ω)) .
kw1 (t) − w2 (t)kL2 (Ω) ≤
π 2T
Proof. Note that w(t) is well-defined because wmn (t) = 0 if m2 + n2 > M . This fact
6
also implies that w = P w. Now for two solutions w1 , w2 we have
kw1 (t) − w2 (t)k2L2 (Ω) =
π2 X
|w1,mn (t) − w2,mn (t)|2
4 m,n≥1
2
T
Z
X
(T −t)(m2 +n2 )
π2
2 +n2 )
(s−t)(m
e
=
(g
−
g
)
−
e
(f
−
f
)
(s)ds
1
2
mn
1
2
mn
4
m,n≥1;m2 +n2 ≤M
t
2
ZT
X
(T −t)M
4
(T −t)Mε
ε
≤ 2
e
kg1 − g2 kL1 (Ω) + e
kf1 (s) − f2 (s)kL1 (Ω) ds
π
m,n≥1;m2 +n2 ≤M t
2
4
≤ 2 Mε e2(T −t)Mε kg1 − g2 kL1 (Ω) + kf1 − f2 kL1 (0,T ;L1 (Ω)) .
π
Here we have used |vmn | ≤ |v|L1 (Ω) and the fact that
#{(m, n) ∈ Z2 |m, n ≥ 1, m2 + n2 ≤ Mε } ≤ Mε .
Reviewing the value of Mε , we have the desired estimate.
Remark 1. A significant convenience of our method is that it is very easy to compute
and represent explicitly the solution w defined by (9). Moreover, this solution is very
smooth because w(t) = P w(t) ∈ C ∞ (Ω) ∩ H01 (Ω) for all t ∈ [0, T ].
Remark 2. The stability magnitude of our well-posed problem is of order
p
t−T
ln(ε−1 )ε 2T .
It is much better, especially when t = 0, than the stability magnitudes given by quasireversibility method and quasi-boundary value method, for example, t−T
T
in [3, 19]
and ( ln(−1 ))−1 in [4, 18].
Our regularized solution is the solution produced directly by the well-posed problem in the previous section from the given data which works even on a very weak
assumption on the exact solution.
Theorem 3 (Regularized solution). Assume that the problem (1 − 3) has at most
one (weak) solution u ∈ C ([0, T ]; L2 (Ω)) ∩ C 1 ((0, T ); L2 (Ω)) corresponding to f ∈
L1 (0, T ; L1 (Ω)) and g ∈ L1 (Ω). Let f and g be measured data satisfying
kf − f kL1 (0,T ;L1 (Ω)) ≤ , kg − gkL1 (Ω) ≤ .
Define the regularized solution u ∈ L1 (0, T ; L2 (Ω)) from f and g as in (9). Then
for each t ∈ [0, T ], u (t) ∈ C ∞ (Ω) ∩ H01 (Ω) and lim u (t) = u(t) in L2 (Ω).
ε→0
7
Proof. We shall use the notations P and M defined in (8). Note that u (t) =
P u (t) ∈ C ∞ (Ω) ∩ H01 (Ω) as in Remark 1. Moreover using the stability in Theorem
2 we find that
kuε (t) − u(t)kL2 (Ω) ≤ kP uε (t) − Pε u(t)kL2 (Ω) + kPε u(t) − u(t)kL2 (Ω)
1/2

p
−1
X
4 ln(ε ) T +t π 
√
≤
ε 2T +
|umn (t)|2  (10)
2
π 2T
2
2
m,n≥1;m +n >Mε
and it must converge to 0 as → 0. To obtain the convergence of the second term in
the right-hand side of (10), we note that
π2 X
|umn (t)|2 = ku(t)k2L2 (Ω) < ∞.
4 m,n≥1
and M → ∞ as → 0.
In the above theorem, we did not give an error estimate because the condition
of the exact solution u is so weak (we even did not require u(t) ∈ H01 (Ω)). However
in practical application we may expect that the exact solution is smoother. In these
cases many explicit errors estimates are available in the next section. An essential
point here is that the regularized solution is the same in any case. This is a substantial
pleasure for practical application because even if someones do not know how good
the exact solution is they are always ensured that the regularized solution works as
well as possible without any further adjustment.
3
Error estimates
From the usual viewpoint from variational method, it is natural to assume that u(t) ∈
H01 (Ω) for all t ∈ [0, T ]. Moreover, if f is smooth and u is a classical solution for the
heat equation (1) then u(t) ∈ H 2 (Ω) ∩ H01 (Ω) for all t ∈ [0, T ]. For these two cases
we have the following explicit error estimates.
Theorem 4 (Error estimate for usual cases). Let u, u as in Theorem 3 and let
t ∈ [0, T ].
(i) Assume that u(t) ∈ H01 (Ω). Then lim u (t) = u(t) in H01 (Ω) and
→0
kuε (t) − u(t)kL2 (Ω)
p
√
4 ln(ε−1 ) T +t
2T
√
≤
ε 2T + p
k∇u(t)kL2 (Ω) .
π 2T
ln(ε−1 )
8
(ii) Assume that u(t) ∈ H 2 (Ω) ∩ H01 (Ω). Then lim u (t) = u(t) in H 2 (Ω) and
→0
p
4 ln(ε−1 ) T +t
2T
√
kuε (t) − u(t)kL2 (Ω) ≤
ku(t)kH 2 (Ω)
ε 2T +
ln(ε−1 )
π 2T
√
2 ln(ε−1 ) T +t
2T
kuε (t) − u(t)kH 1 (Ω) ≤
ε 2T + p
ku(t)k2H 2 (Ω) .
0
−1
πT
ln(ε )
Here we use the norm
kwk2H 1 = k∇wk2L2 = kwx k2L2 + kwy k2L2 ,
0
= kwk2L2 + kwk2H 1 + kwxx k2L2 + kwxy k2L2 + kwyx k2L2 + kwyy k2L2 .
kwk2H 2
0
Proof. (i) By using the integral by part and the Parseval equality, it is straightforward
to check that if u(t) ∈ H01 (Ω) then
π2 X
(m2 + n2 )|umn (t)|2 = k∇u(t)k2L2 (Ω) .
4 m,n≥1
(11)
Using (11) we have
X
4
1 X
k∇u(t)k2L2 (Ω) .
(m2 + n2 )|umn (t)|2 = 2
|umn (t)|2 ≤
M
π
M
ε m,n≥1
ε
2
2
m,n≥1;m +n >Mε
Substituting the latter inequality into the estimate (10) in the proof of Theorem 3,
we obtain the error estimate in L2 .
To prove the convergence in H01 we use the identity (11) and the stability of
Theorem 2 again
k∇uε (t) −
=
π2
4
+
≤
+
X
(m2 + n2 )|umn (t)|2
m,n≥1;m2 +n2 >Mε
X
Mε |(Pε uε )mn (t) − (Pε u)mn (t)|2
m,n≥1;m2 +n2 ≤Mε
2
X
π
4
π2 X
=
(m2 + n2 )|uε,mn (t) − umn (t)|2
4 m,n≥1
(m2 + n2 )|uε,mn (t) − umn (t)|2
m,n≥1;m2 +n2 ≤Mε
2
X
π
4
π2
4
∇u(t)k2L2 (Ω)
(m2 + n2 )|umn (t)|2
m,n≥1;m2 +n2 >Mε
≤ Mε kPε uε (t) − Pε u(t)k2L2 (Ω) +
≤
4(ln(ε−1 ))2 T +t π 2
ε T +
π2T 2
4
π2
4
X
m,n≥1;m2 +n2 >Mε
X
m,n≥1;m2 +n2 >Mε
9
(m2 + n2 )|umn (t)|2
(m2 + n2 )|umn (t)|2 .
(12)
The second term in the right-hand side in (12) converges to 0 as → 0 because the
convergence in (11). Thus the convergence in H01 has been proved.
(ii) We now assume that u(t) ∈ H 2 (Ω) ∩ H01 (Ω). We have an identity similar to (11)
π2 X
(m2 + n2 )2 |umn (t)|2
4 m,n≥1
= kuxx (t)k2L2 (Ω) + kuxy (t)k2L2 (Ω) + kuyx (t)k2L2 (Ω) + kuyy (t)k2L2 (Ω) .
(13)
The error estimate in L2 (Ω) follows (10) and the following inequality
X
|umn (t)|2 ≤
m,n≥1;m2 +n2 >Mε
1 X
4
(m2 + n2 )2 |umn (t)|2 ≤ 2 2 ku(t)k2H 2 (Ω) .
2
Mε m,n≥1
π Mε
Similarly, from (12) and the estimate
X
(m2 + n2 )|umn (t)|2 ≤
m,n≥1;m2 +n2 >Mε
≤
1 X
(m2 + n2 )2 |umn (t)|2
Mε m,n≥1
4
π 2 Mε
ku(t)k2H 2 (Ω) ,
we find that
π 2 (ln(ε−1 ))2 T +t
1
T
ε
+
ku(t)k2H 2 (Ω) .
2
4T
Mε
√
√
Using the inequality a + b ≤ ( a + b)2 we obtain the error estimate in H01 .
k∇uε (t) − ∇u(t)k2L2 (Ω) ≤
Finally we prove the convergence in H 2 (Ω). Similarly to (12) we have
k(uε − u)xx (t)k2L2 + k(uε − u)xy (t)k2L2 + k(uε − u)yx (t)k2L2 + k(uε − u)yy (t)k2L2
π2 X
=
(m2 + n2 )2 |uε,mn (t) − umn (t)|2
4 m,n≥1
≤
π2
4
X
Mε2 |uε,mn (t) − umn (t)|2
m,n≥1;m2 +n2 ≤Mε
2
X
+
π
4
(m2 + n2 )2 |umn (t)|2
m,n≥1;m2 +n2 >Mε
≤ Mε2 kPε uε (t) − Pε u(t)k2L2 (Ω) +
−1
≤
3
2
2(ln(ε )) T +t π
ε T +
π2T 3
4
π2
4
X
(m2 + n2 )2 |umn (t)|2
m,n≥1;m2 +n2 >Mε
X
(m2 + n2 )2 |umn (t)|2 → 0
m,n≥1;m2 +n2 >Mε
as → 0 due to the convergence in (13).
10
(14)
Remark 3. In Theorem 4 we have pointwise estimates due to the pointwise condition
on the exact solution u. As a consequence, we shall immediately obtain a uniform
convergence whenever the corresponding uniform condition is imposed. For example,
if the exact solution u is in C([0, T ]; H01 (Ω)) or C([0, T ]; H 2 (Ω)∩H01 (Ω)) then we have
the estimates kuε − ukC([0,T ];L2 (Ω)) and kuε − ukC([0,T ];H 1 (Ω)) , respectively, in the same
0
form of estimates in Theorem 4.
Remark 4. The error estimates in Theorem 4 work well no matter t > 0 or t = 0.
In many earlier works, we find that the error ku (0) − u(0)kL2 (Ω) is often not given
(e.g. [17]) and an explicit error estimate in H01 (Ω) is not available (e.g. [3, 7, 4, 17,
19, 18]).
In Theorem 4 (ii), an error estimate in H 2 (Ω) is not given because we do not
have enough information on the exact solution (we just know u(t) ∈ H 2 (Ω) ∩ H01 (Ω)).
However, when u is smoother then an explicit estimate in H 2 (Ω) may be derived.
In the last theorem, we shall give the error estimates in some special cases when
the exact solution is very good. We see from the proof of Theorem 4 that the facts
u(t) ∈ H01 (Ω) and u(t) ∈ H 2 (Ω) ∩ H01 (Ω) are equivalent to
X
(m2 + n2 )k |umn (t)|2 < ∞
m,n≥1
with k = 1, 2, respectively. We shall see that from the latter condition with k > 2
we may improve the estimate, and in particular give an error estimate in H 2 (Ω). We
next consider a stronger condition similar to (in fact, weaker than)
sup
X
e2T (m
2 +n2 )
|umn (t)|2 < ∞
(15)
t∈[0,T ] m,n≥1
which is a two-dimensional version of the condition (4) in [14]. Such a condition seems
essential to solve the nonlinear problem. Although it is quite strict for the linear case,
as we discussed in the first section, if the above condition (15) holds then we have a
√
very good convergence rate which is of order (ln(−1 ))3/2 .
Theorem 5 (Error estimate for special cases). Let u, u as in Theorem 3 and let
t ∈ [0, T ]. (i) Assume that
Ek (t) =
∞
X
(n2 + m2 )k u2mn (t) < ∞
n,m=1
11
(16)
for some constant k > 2. Then
kuε (t) − u(t)kL2 (Ω)
kuε (t) − u(t)kH 1 (Ω)
0
kuε (t) − u(t)kH 2 (Ω)
p
p
k2
4 ln(ε−1 ) T +t π Ek (t)
2T
2T
√
≤
ε
+
2
ln(ε−1 )
π 2T
p
k−1
2
2T
2 ln(ε−1 ) T +t π Ek (t)
ε 2T +
≤
πT
2
ln(ε−1 )
p
k−2
2
2T
2(ln(ε−1 ))3/2 T +t 3π Ek (t)
√
ε 2T +
.
≤
2
ln(ε−1 )
πT T
Here we assume ≤ e−2T for the estimate in H 2 (Ω).
(ii) Assume that
Fr (t) =
X
2 +n2 )
e2r(m
|umn (t)|2 < ∞
m,n≥1
for some constant r > 0. Then
p
p
4 ln(ε−1 ) T +t π Fr (t) r
√
≤
ε 2T
ε 2T +
2
π 2T
p
2 ln(ε−1 ) T +t π Fr (t) ln(ε−1 ) r
√
ε 2T +
ε 2T
≤
πT
2 2T
p
6(ln(ε−1 ))3/2 T +t 3π Fr (t) ln(ε−1 ) r
√
≤
ε 2T +
ε 2T .
4T
πT T
kuε (t) − u(t)kL2 (Ω)
kuε (t) − u(t)kH 1 (Ω)
0
kuε (t) − u(t)kH 2 (Ω)
Here we assume ≤ e−2T for the estimate in H01 (Ω), and ≤ e−4T for the estimate
in H 2 (Ω).
Proof. (i) We use the same way of the proof of Theorem 4. We shall prove the error
estimates in H 2 (Ω) (the other ones are similar and easier). From
X
(m2 + n2 )2 |umn (t)|2 ≤
m,n≥1;m2 +n2 >Mε
1
X
Mεk−2 m,n≥1
(m2 + n2 )k |umn (t)|2 ≤
Ek (t)
Mεk−2
and (14) we find that
k(uε − u)xx (t)k2L2 + k(uε − u)xy (t)k2L2 + k(uε − u)yx (t)k2L2 + k(uε − u)yy (t)k2L2
!2
p
2(ln(ε−1 ))3 T +t
4 3/2 T +t π Ek (t) − k−2
π 2 Ek
≤
.
ε T +
≤
M ε 2T +
Mε 2
π2T 3
4Mεk−2
π ε
2
Using
kwkH 2 ≤ kwkL2 + kwkH 1
0
q
+ kwxx k2L2 + kwxy k2L2 + kwyx k2L2 + kwyy k2L2
12
(17)
and M ≥ 1 we conclude the desired estimate in H 2 (Ω).
(ii) From (10) and
X
X
|umn (t)|2 ≤ e−2rMε
m,n≥1;m2 +n2 >Mε
e2r(m
2 +n2 )
r
|umn (t)|2 ≤ Fr (t)ε T
m,n≥1
we get the error estimate in L2 (Ω).
Note that the function ξ 7→ eξ /ξ is increasing when ξ ≥ 1. Thus
(m2 + n2 ) ≤ Mε e2r(m
2 +n2 −M
ε)
when m2 + n2 > Mε ≥ 1.
It implies that
X
m,n≥1;m2 +n2 >Mε
2 +n2 −M
X
(m2 + n2 )|umn (t)|2 ≤ Mε
e2r(m
ε)
r
|umn (t)|2 ≤ Mε Fr (t)ε . T
m,n≥1
The error estimate in H01 (Ω) follows the above estimate and (12).
Similarly, because the function ξ 7→ eξ /ξ 2 is increasing when ξ ≥ 2, we find that
(m2 + n2 )2 ≤ Mε2 e2r(m
2 +n2 −M
ε)
if m2 + n2 > Mε ≥ 2.
It follows that
X
m,n≥1;m2 +n2 >Mε
2 +n2 −M )
ε
X
(m2 + n2 )|umn (t)|2 ≤ Mε2
e2r(m
r
|umn (t)|2 ≤ Mε2 Fr (t)ε T .
m,n≥1
Thus (14) reduces to
k(uε − u)xx (t)k2L2 + k(uε − u)xy (t)k2L2 + k(uε − u)yx (t)k2L2 + k(uε − u)yy (t)k2L2
!2
p
π
F
(t)
T +t
r
r
2(ln(ε−1 ))3 T +t
4
r
≤
ε T + Mε2 Fr (t)ε T ≤
M 3/2 ε 2T +
Mε ε 2T
.
π2T 3
π ε
2
Using (17) again and M ≥ 1 we conclude the error estimate in H 2 (Ω).
Remark 5. If (15) holds, i.e.
P
e2T (m
2 +n2 )
|umn (t)|2 < ∞, then applying Theorem
m,n≥1
5 in the case r = T we get
√
sup kuε (t) − u(t)kL2 (Ω) ≤ C ε,
t∈[0,T ]
√
sup kuε (t) − u(t)kH 1 (Ω) ≤ C ln(ε−1 ) ε,
t∈[0,T ]
0
√
sup kuε (t) − u(t)kH 2 (Ω) ≤ C(ln(ε−1 ))3/2 ε.
t∈[0,T ]
13
Notice that in [11], under a similar condition, Liu gave the error estimate (See Theorem 3.3, page 466)
t0
1
kgα − g0 kL2 (Ω) ≤ C( 2 )1− T
Let t0 = 0, we get
kgα
− g0 kL2 (Ω)
p 1
≤ C 2
Thus at t = 0, our method gave the same order error in L2 -norm as the method of Liu
[11]. However, the strong point of our paper is that the error estimates in H01 (Ω) or
H 2 (Ω) established, and also of Hölder type (in fact, they are better than any estimate
of order q with q ∈ (0, 1/2)). They are not given in [11].
Remark 6. The truncated regularization method is a very simple and effective method
for solving some ill-posed problems and it has been successfully applied to some inverse
heat conduction problems [5, 8, 13]. Recently, in [14] many applications for a model
of the Helmholtz equation are introduced and a Fourier method was applied for solving
a Cauchy problem for the Helmholtz equation. In [9], ChuLiFu and his group used the
truncated method to solve the backward heat in the unbounded region and established
the logarithimic order of the form
ku(., t) − uδ,ξmax k
t
≤ E 1− T
ln
E
δ
)s
(t−T
2T

1 +
ln
1
T
ln
E
δ
! 2s 
E
δ
+ ln(ln
E −s
) 2T
δ
.
(18)
And in [20], the authors gave the following estimates
kwβ,aβ (., ., t) − u(., ., t)k
1 −α(T −t) 1 α
≤ β t/T (ln ) 2T
exp(k 2 (T − t)2 ) + Q(β, t, u)(ln ) 2 .
β
β
(19)
And in [21], Trong and Tuan only established the logarithmic form as follows
ku(., ., t) − u (., ., t)k ≤
C
1 + ln( T )
(20)
Note that the errors (18), (19) and (20) are the same order as Theorem 5 (i).
However,the logarithmic type estimate is, in general, much worse than any Hölder
type estimate, i.e. q for some q > 0. In Theorem 5 (ii) we also establish this type of
estimates, which are not given in [9, 20, 21]. It worth mentioning that our regularized
solution is unique, in all cases. This proves that our method is effective.
14
Remark 7. Sometime, it is also important to considerd the 2-D backward heat for
a general two-dimensional domain, e.g. [12]. In this case to apply the truncation
method, we need to consider the spectral problem of operator −∆ in this domain (with
homogeneous Dirichlet boundary condition). However, this question is not always
solvable explicitly and this is a disadvantage point of our method.
4
Numerical experiments
In this section we give some numerical experiments for our method. For simplicity,
we shall recover the initial temperature at t = 0 from the final data at T = 1.
Example 1. Consider the problem
ut − ∆u = f (x, y, t) ≡ 3et sin(x) sin(y),
with the final condition
u(x, y, 1) = g(x, y) ≡ e sin(x) sin(y).
Problem (1)-(3) with exact data (f, g) has the exact solution
u(x, y, t) = et sin(x) sin(y).
For any n = 1, 2, ..., let us take the measured data
fn = f, gn (x, y) = g(x, y) + n sin(nx) sin(ny).
Then Problem (1)-(3) with measured data (f , g ) has corresponding solution
u
en (x, y, t) = u(x, y, t) +
1 n2 (1−t)
e
sin(nx) sin(ny).
n
We see that
4
→ 0,
n
2
en
=
→ +∞
n
kgn (x, y) − g(x, y)kL1 (Ω) =
ke
un (., ., 0) − u(., ., 0)kL2 (Ω)
It means that if n is large then a small error of data might cause a large error of
solutions. Therefore, the problem is really unstable and hence a regularization is
15
necessary. Using the regularization of Theorem 2 correspoding ε = 4/n, we see that
when n > 4e4 then the regularized solution at t = 0 is
uε (x, y) = sin(x) sin(y),
which coincides the exact solution u(., ., 0). In this example, our method works very
well because the exact solution’s form is of a truncated Fourier series.
Example 2. Consider the problem
ut − ∆u = f (x, y, t) ≡ sin(x) ty 3 − πty 2 − 6ty + 6y + 2tπ − 2π
with the final condition
u(x, y, 1) = g(x, y) ≡ 0
The exact solution of the latter equation is
u(x, y, t) = (1 − t) sin(x)y 2 (π − y).
For any n = 1, 2, ..., take the measured data
fn = f, gn (x, y) =
1
sin(nx) sin(ny).
4n
Then the disturbed solution is
u
en (x, y, t) = u(x, y, t) +
1 n2 (1−t)
e
sin(nx) sin(ny).
4n
We see that
1
→ 0,
n
2
en
=
→ +∞.
4n
kgn (x, y) − g(x, y)kL1 (Ω) =
ke
un (., ., 0) − u(., ., 0)kL2 (Ω)
Thus the problem in this case is also unstable. We now compute the regularized
solutions by using the regularization method introduced in the previous sections with
ε = 1/n. The effect of our regularization is represented via Table 1 below, where we
denote by u := u (., ., 0) the regularized value at t = 0, u
e := u
en (., ., 0) the disturbed
value (with = 1/n), and u0 := u(., ., 0) the exact value. We can see that while the
errors between the disturbed solution and the exact solution is extremely large, the
error between the regularized solution and the exact solution is acceptable, even in
H01 -norm.
16
1
n
−3
ε=
uε
ke
u − u0 kL2
kuε − u0 kL2
0
10
4 sin(x) sin(y)
10−5
4 sin(x) sin(y) − 32 sin(x) sin(2y) exp(1010 )
0.391677
1.600311
10−9
4 sin(x) sin(y) − 32 sin(x) sin(2y) exp(1018 )
0.315050
1.421075
0.111858
0.738103
+
10−15
4
27
exp(999992) 2.388527
kuε − u0 kH 1
sin(x) sin(3y)
4 sin(x) sin(y) − 32 sin(x) sin(2y) exp(1030 )
+
4
27
5.506293
sin(x) sin(3y)
3
sin(x) sin(4y)
− 16
Table 1. Errors between disturbed solution, regularized solution and exact solution.
Acknowledgments
Most part of the work was done when the first author was a student of University of
Science, Vietnam National University at HoChiMinh City. The second and the third
authors are supported by the Council for Natural Sciences of Vietnam. We thank the
referees for constructive comments leading to the improved version of the paper.
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19